Answer to Question #132028 in Real Analysis for Joel George

Question

Expand f(x) = cos x, 0 < x < π/2 , in a Fourier sine series.

Answer 1

Assume that f(x) is an odd function on the interval [−L/2, L/2].

$$a_m=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\cos(\dfrac{m\pi x}{L})dx=0, m=0,1,2,...$$

$$b_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\sin(\dfrac{n\pi x}{L})dx, n=1,2,...$$

$$b_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\sin(\dfrac{n\pi x}{L})dx=$$

$$=\dfrac{2}{\pi/2}\displaystyle\int_{0}^{\pi/2}\cos x\sin(\dfrac{n\pi x}{\pi/2})dx=$$

$$=\dfrac{4}{\pi}\displaystyle\int_{0}^{\pi/2}\cos x\sin(2nx)dx$$

$$\int\cos x\sin (2nx)dx=$$

$$=\dfrac{1}{2}\int(\sin ((2n+1)x)+\sin ((2n-1)x))dx=$$

$$=-\dfrac{1}{2(2n+1)}\cos((2n+1)x)-\dfrac{1}{2(2n-1)}\cos((2n-1)x)+C$$

$$b_n=\dfrac{4}{\pi}\displaystyle\int_{0}^{\pi/2}\cos x\sin(2nx)dx=$$

$$=-\dfrac{2}{\pi(2n+1)}(\cos((2n+1)\dfrac{\pi}{2})+\dfrac{2}{\pi(2n+1)}\cos((2n+1)(0))-$$

$-\dfrac{2}{\pi(2n-1)}(\cos((2n-1)\dfrac{\pi}{2})+\dfrac{2}{\pi(2n-1)}\cos((2n-1)(0))=$

$$=\dfrac{2}{\pi(2n+1)}\sin(\pi n)+\dfrac{2}{\pi(2n+1)}+$$

$$+\dfrac{2}{\pi(2n-1)}\sin(\pi n)+\dfrac{2}{\pi(2n-1)}=\dfrac{8n}{\pi(4n^2-1)}$$

$$f(x)\sim\displaystyle\sum_{n=1}^\infin\dfrac{8n}{\pi(4n^2-1)}\sin(2nx)$$